The given figure represents the lines y = x + 1 and y = $\sqrt{3}x - 1.$ Write down the angles which the lines make with the positive direction of the x-axis. Hence, determine θ.

Given,
y = x + 1 and y = $\sqrt{3}x - 1$.

Comparing equations with y = mx + c we get,

m₁ = 1 and m₂ = $\sqrt{3}$.

Let the first line make angle θ₁ and second make θ₂ with positive direction of x-axis.

The inclination that y = x + 1 makes is,
⇒ m₁ = tan θ₁ = 1
⇒ tan θ₁ = 1 = tan 45°
⇒ tan θ₁ = tan 45°
⇒ θ₁ = 45°.

The inclination that y = $\sqrt{3}x$ - 1 makes is,
⇒ m₂ = tan θ₂ = $\sqrt{3}$
⇒ tan θ₂ = $\sqrt{3}$ = tan 60°
⇒ tan θ₂ = tan 60°
⇒ θ₂ = 60°.

From graph we get,

60° is the exterior angle. We know that,

Exterior angle = Sum of two opposite interior angles.

∴ 60° = θ + 45°
⇒ θ = 60° - 45°
⇒ θ = 15°

Hence, y = x + 1 makes 45° and y = $\sqrt{3}x - 1$ makes 60° with the x-axis. The value of θ = 15°.